16-Civ-B1 Advanced Structural Analysis · December 2019
Question 7 of 9: Flexibility analysis of a non-prismatic fixed-ended beam
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examination, December 2019 —
16-Civ-B1 Advanced Structural Analysis, 3 hours, CLOSED BOOK (an
approved Casio or Sharp calculator only). Nine questions on seven pages:
Questions 1 and 2 are compulsory (8 and 12 marks); the candidate then answers
TWO of Questions 3, 4 or 5 (18 marks each) and TWO of Questions 6, 7, 8 or 9
(22 marks each), so six questions make a complete paper of 100 marks. Because
this set is a study resource, all nine questions are solved
here.
Reference texts. R. C. Hibbeler, Structural
Analysis, 10th ed. (Pearson) — Ch. 6 influence lines, Ch. 8–9
deflections and virtual work, Ch. 10 force (flexibility) method, Ch. 11
slope-deflection, Ch. 12 moment distribution, Ch. 15–16 matrix stiffness.
A. Kassimali, Structural Analysis, 6th ed. (Cengage) — Ch. 7
Castigliano and least work, Ch. 8 influence lines, Ch. 13 slope-deflection,
Ch. 17–18 matrix stiffness. A. Ghali, A. M. Neville and T. G. Brown,
Structural Analysis: A Unified Classical and Matrix Approach, 7th ed.
(CRC). Canadian context for the settlement question: NBC 2020 Part 4 and
CSA S6:19 treat differential support settlement as an imposed deformation to be
combined with the permanent loads, so the self-equilibrating moment set computed
in Question 6 is a real design action, not a curiosity.
Sign convention used throughout.
End moments are written counter-clockwise-positive on the member, which is the
convention that matches the standard six-degree-of-freedom element stiffness
matrix; the slope-deflection equation is then
$M_{ij} = \dfrac{2EI}{L}\left(2\theta_i + \theta_j - 3\psi_{ij}\right) +
\mathrm{FEM}_{ij}$ with $\psi$ the counter-clockwise chord rotation and, for a
downward UDL, $\mathrm{FEM}_{ij} = +wL^2/12$, $\mathrm{FEM}_{ji} = -wL^2/12$.
Ordinary sagging bending moments, which is what every diagram below
plots, are recovered as $M_{\text{sag}}(i) = -M_{ij}$ and
$M_{\text{sag}}(j) = +M_{ji}$. Mixing this with the clockwise-positive
convention of some textbooks produces clean-looking integers that are wrong.
Question 7: Flexibility analysis of a non-prismatic fixed-ended beam
(22 marks)
Two 20 kN point loads act downwards at $x = 3\ \text{m}$ and
$x = 9\ \text{m}$; total span 12 m; both ends built in. The structure and the
loading are symmetric about mid-span.
Find. The fixed-end moments $M_A$ and $M_B$.
Question 7: fixed-ended beam of 2EI / EI / 2EI over 3 m / 6 m / 3 m carrying two 20 kN loads.
Approach. Release both end moments to leave a simply
supported primary structure, note that symmetry makes the two redundants equal
so their combined effect is a constant added to the primary moment
diagram, and impose the single compatibility condition that the total
$M/EI$ area vanishes.
Choose the primary structure and the redundants. The beam
is three times statically indeterminate, but with no horizontal load the axial
redundant is idle, so two redundants remain: the end moments $X$ at A and at B.
Releasing both leaves a simply supported span of 12 m — the primary
structure. Because both the geometry and the loading are symmetric about
mid-span, the two redundants are equal, and a pair of equal hogging end moments
of magnitude $X$ produces a uniform moment $-X$ along the whole beam.
Hence
$$M(x) = M_0(x) - X$$
where $M_0$ is the simply supported moment diagram.
Write the primary moment diagram. Each reaction of the
simply supported beam is 20 kN, so
$$M_0(x) = \begin{cases}
20x & 0 \le x \le 3\\
60 & 3 \le x \le 9\\
20(12-x) & 9 \le x \le 12
\end{cases}$$
with a flat 60 kN·m plateau between the loads, peaking at
60 kN·m.
State the compatibility condition. Both ends are built in,
so the relative rotation between them is zero:
$$\theta_B - \theta_A = \int_0^{L}\frac{M(x)}{EI(x)}\,\mathrm{d}x = 0
\quad\Longrightarrow\quad
\int_0^{L}\frac{M_0(x)}{EI(x)}\,\mathrm{d}x
= X\int_0^{L}\frac{\mathrm{d}x}{EI(x)} .$$
In force-method notation these two integrals are $\delta_{10}$ and $f_{11}$ for
the redundant pair with $\bar m = 1$ throughout — the constant unit-moment
field produced by the equal end moments. Symmetry supplies the second condition
automatically ($\theta_A = -\theta_B$, so each is separately zero).
Evaluate $\delta_{10}$, segment by segment. Working in
units of $1/EI$:
$$\int_0^{3}\frac{20x}{2}\,\mathrm{d}x = \frac{90}{2} = 45.0,\qquad
\int_3^{9}\frac{60}{1}\,\mathrm{d}x = 360.0,\qquad
\int_9^{12}\frac{20(12-x)}{2}\,\mathrm{d}x = 45.0$$
$$\delta_{10} = 45.0 + 360.0 + 45.0 = \frac{450.0}{EI}$$
Solve for the redundant.
$$X = \frac{\delta_{10}}{f_{11}} = \frac{450.0}{9.0}
= \boxed{M_A = M_B = 50.0\ \text{kN}\cdot\text{m}\ \ (\text{hogging})}$$
The $EI$ cancels, as it must: only the relative stiffnesses matter,
which is why the question supplies them as ratios and no absolute value at
all.
Complete the diagram and check. Subtracting the constant,
the moment under each load is $60.0 - 50.0 = +10.0\ \text{kN}\cdot\text{m}$
sagging, and the whole middle six metres carries that same value because $M_0$
is flat there. The points of contraflexure sit where $20x = 50$, that is
$x = 2.50\ \text{m}$ from each end — inside the stiff haunches, not at the
loads. Statics closes: the two reactions are 20 kN each and
$20(3) - 50.0 = 10.0$.
Question 7: the final bending moment diagram is the simply supported diagram shifted down by the constant redundant, 50.0 kN.m.
Compare with the prismatic beam. If the haunches were
removed and $EI$ were uniform, the same integrals give
$X = \left(\int M_0\,\mathrm{d}x\right)/L = 540.0/12 = 45.0\ \text{kN}\cdot\text{m}$,
so the haunches attract about 11 per cent more moment to the supports and
correspondingly relieve mid-span from 15.0 to 10.0 kN·m. Stiffness
attracts moment: that one sentence is the whole engineering content of the
question, and quoting the prismatic value alongside the answer is the cheapest
way to show it has been understood.
